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    (Original post by ZJuwelH)
    I get the gist of this but the step in bold, where's it from and what's it mean?
    10a + b = 11(a + b)

    10a + b = 11a + 11b

    11a + 11b - 10a -b = 0

    11a - 10a + 11b - b = 0

    a + 10b = 0

    (it means either a and b are both 0, or a and b are of opposite sign, which cannot be.).
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    (Original post by elpaw)
    10a + b = 11(a + b)

    10a + b = 11a + 11b

    11a + 11b - 10a -b = 0

    11a - 10a + 11b - b = 0

    a + 10b = 0

    (it means either a and b are both 0, or a and b are of opposite sign, which cannot be.).
    Duuuh <slaps forehead> and I want to study Maths at Cambridge...

    But the one-digit number 0 would work
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    (Original post by ZJuwelH)
    Duuuh <slaps forehead> and I want to study Maths at Cambridge...

    But the one-digit number 0 would work
    Show your worth and finish the problem off
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    (Original post by ZJuwelH)
    But the one-digit number 0 would work
    yes it would. but it's a trivial answer. (as certain lecturers say).
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    (Original post by theone)
    You have to start by proving you can only do this when the number in question is a 3 digit number. Then let n (the 3 digit number) = 100a + 10b + c where a,b,c are integers <10 and then set that equal to 11(a+b+c) (i.e. 11 times the sum of their digits)
    Let me know if you're stuck again.
    n = 100a+10b+c = 11(a+b+c)
    100a+10b+c = 11a+11b+11c
    89a = b+10c (the most constructive rearrangement of this equation I could find, and even then I'm stuck)...

    Sorry theone I flopped your challenge, looks like it's Manchester for me!
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    (Original post by ZJuwelH)
    n = 100a+10b+c = 11(a+b+c)
    100a+10b+c = 11a+11b+11c
    89a = b+10c (the most constructive rearrangement of this equation I could find, and even then I'm stuck)...

    Sorry theone I flopped your challenge, looks like it's Manchester for me!
    there is a special condition that links a and c, but i've forgotten it. i think this question was in the STEP 1 paper i took in the summer.
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    (Original post by elpaw)
    there is a special condition that links a and c, but i've forgotten it. i think this question was in the STEP 1 paper i took in the summer.
    Well it's clear a can only be 1 and then b and c must possess a certain value?

    What about numbers greater than 3 digits...
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    (Original post by theone)
    Well it's clear a can only be 1 and then b and c must possess a certain value?

    What about numbers greater than 3 digits...
    Sorry it's not clear to me why a = 1, is it because otherwise the number isn't three digits? Not at my mental best, bloody Chelsea how dare they :mad:
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    (Original post by ZJuwelH)
    Sorry it's not clear to me why a = 1, is it because otherwise the number isn't three digits? Not at my mental best, bloody Chelsea how dare they :mad:
    We have 89a = 10b + c agreed?

    Now b and c <10 so the RHS < 100 so 89a < 100 so a must be 1.
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    (Original post by theone)
    We have 89a = 10b + c agreed?

    Now b and c <10 so the RHS < 100 so 89a < 100 so a must be 1.
    Duuuh again. At this rate I won't even get in to Manchester, it's Queen Mary for me! Still, I'm going to try this,...
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    (Original post by ZJuwelH)
    Duuuh again. At this rate I won't even get in to Manchester, it's Queen Mary for me! Still, I'm going to try this,...
    since that is the case, then c must be 8. And b must be 9.
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    So if 89a = 10b+c and a = 1:

    89 = 10b + c
    But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

    So far I have 0 and 189 yay!
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    (Original post by 2776)
    since that is the case, then c must be 8. And b must be 9.
    so 198 = 11 (1 + 9 + 8) = 198 !

    but there could be other (4+ digit) answers....
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    (Original post by ZJuwelH)
    So if 89a = 10b+c and a = 1:

    89 = 10b + c
    But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

    So far I have 0 and 189 yay!
    is 0 an interger?
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    (Original post by ZJuwelH)
    So if 89a = 10b+c and a = 1:

    89 = 10b + c
    But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

    So far I have 0 and 189 yay!
    *cough* 198 *cough*
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    (Original post by ZJuwelH)
    So if 89a = 10b+c and a = 1:

    89 = 10b + c
    But since b and c are integers less than 10 the only combination that works is b = 8 and c = 9 so n = 189...

    So far I have 0 and 189 yay!
    Doh! It's wrong!
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    (Original post by elpaw)
    so 198 = 11 (1 + 9 + 8) = 198 !

    but there could be other (4+ digit) answers....
    But there arn't. rep to the first person to prove why
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    (Original post by 2776)
    is 0 an interger?
    yes
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    (Original post by ZJuwelH)
    Doh! It's wrong!
    you're not going to even get into QM
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    (Original post by elpaw)
    so 198 = 11 (1 + 9 + 8) = 198 !

    but there could be other (4+ digit) answers....
    198 is the only one

    4 digit numbers gives:

    989a + 89b = 10d + c

    Which is impossible for a,b,c,d < 10 because max of 10d + c is 99 but the minimum of the lhs is 989 because a has to be at least 1
 
 
 
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